\documentclass[reqno]{amsart}
\usepackage{hyperref}
\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2010(2010), No. 38, pp. 1--9.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2010 Texas State University - San Marcos.}
\vspace{9mm}}
\begin{document}
\title[\hfilneg EJDE-2010/38\hfil Positive solutions]
{Positive solutions for second-order singular
three-point boundary-value problems with sign-changing
nonlinearities}
\author[C. Ji, B. Yan\hfil EJDE-2010/38\hfilneg]
{Caisheng Ji, Baoqiang Yan} % in alphabetical order
\address{Caisheng Ji \newline
Department of Mathematics, Shandong Normal University,
Jinan 250014, China}
\email{jicaisheng@163.com}
\address{Baoqiang Yan \newline
Department of Mathematics, Shandong Normal University,
Jinan 250014, China}
\email{yanbqcn@yahoo.com}
\thanks{Submitted July 30, 2009. Published March 14, 2010.}
\thanks{Supported by grants 10871120 from the fund of National
Natural Science, J07WH08 from \hfill\break\indent the
Shandong Education Committee, and Y2008A06 from the Shandong
Natural Science}
\subjclass[2000]{34B10, 34B15}
\keywords{Singular three-point boundary-value problem; sign-changing;
\hfill\break\indent nonlinearity; positive solution; fixed point}
\begin{abstract}
In this article, we study the existence and uniqueness
of the positive solution for a second-order singular
three-point boundary-value problem with sign-changing
nonlinearities. Our main tool is a fixed-point theorem.
\end{abstract}
\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{corollary}[theorem]{Corollary}
\section{Introduction}
In this article, we consider the second-order boundary-value problem
\begin{gather}
x''(t)+f(t,x(t))=0,\quad 00$.
Moreover, when $g(t)$ is a sign-changing function in $[0,1]$
and $f$ is nondecreasing and without any singular points,
using the fixed point theorem of strict-set-contractions,
Bing Liu \cite{l1} established the existence of at least two positive
solutions for \eqref{e1.3}. When $g(t)>0$ and $f$ is a
given sign-changing function without any singular points and
any monotonicity, using the increasing operator theory and
approximation process, Xian Xu \cite{x1} showed at least three solutions
for the three-point boundary-value problem \eqref{e1.3}.
In addition, the existence of solutions of nonlinear multi-point
boundary-value problems have been studied by many other authors;
the readers are referred to \cite{l1,m1,x2,x3}
and the references therein.
Motivated by \cite{h1,y1}, the purpose of this article is to examine
the existence and the uniqueness of the positive solution of
\eqref{e1.1}-\eqref{e1.2} under the assumption that $f$ may be
singular at $t=0$, $x=0$ and be superlinear at $x=\infty$
and change sign.
There are only a few papers considering
\eqref{e1.1}-\eqref{e1.2} under this assumptions.
We try to fill this gap in the literature with this paper.
In this article, we use the following assumptions:
\begin{itemize}
\item[(H1)] $f(t,x)\in C((0,1]\times(0,+\infty),(-\infty,+\infty))$,
\item[(H2)] $k(t), a(t), b(t)\in C((0,1],(0,+\infty))$,
$tk(t)\in L(0,1]$,
\item[(H3)] there exist $F(x)\in C((0,+\infty), (0,+\infty))$,
$G(x)\in C([0,+\infty),[0,+\infty))$ such that
$f(t,x)\leq k(t)(F(x)+G(x))$.
\item[(S1)] $f(t,x)\geq a(t)$ hold for $01$, such that
$\int_1^R\frac{dy}{F(y)}\cdot (1+\frac{\bar{G}(R)}{F(R)})^{-1}
>\int_0^1 sk(s)ds$, where
$\bar{G}(R)=\max_{s\in [0,R]} G(s)$.
\end{itemize}
This paper is organized as follows. In Section 2, we give some
preliminaries. In Section 3, we obtain the existence of at least
one positive solution for \eqref{e1.1}-\eqref{e1.2}, and show an
application of our results.
\section{Preliminaries}
\begin{lemma}[\cite{d1}] \label{lem2.1}
Let $E$ be a Banach space, $R>0$, $B_R=\{x\in E:\|x\|\leq R\}$,
$F:B_R\to E$ be a completely continuous operator.
If $x\neq \lambda F(x)$ for any $x\in E$ with $\|x\|=R$ and
$0[\frac{1}{\eta}+1]$ be a natural number,
$d_n=\min\{b(t):t\in[\frac{1}{n},1]\}$,
$b_n=\min\{d_n,\frac{1}{n}\}$, $C_n=\{x: x\in C[\frac{1}{n},1]\}$
with norm $ \|x\|=\max\{|x(t)|, \frac{1}{n}\leq t\leq 1\}$. It is
easy to see that ($C_n, \| \cdot \|$) is a Banach space.
Inspired by \cite{y1}, we define $T_n$ as
$$
(T_nx)(t)=b_n+\int_{\frac{1}{n}}^1 G_{\frac{1}{n},1}(t,s)
f(s,\max\{b_n,x(s)\})ds,\quad x\in C_n,\; t\in[\frac{1}{n},1],
$$
where
\[
G_{\frac{1}{n},1}(t,s)=\begin{cases}
G_1(t,s), &\frac{1}{n}x(1)$, we can get
a point $t_1\in (\frac{1}{n},\eta)$ such that $x(t_1)=x(1)$.
Let $\gamma=\sup \{t_1:\ t_1\in (\frac{1}{n},\eta), x(t_1)=x(1)\}$.
It follows that $x(\gamma)=x(1)$ and $x(t)(1+\frac{\bar{G}(R)}{F(R)})\int_0^1sk(s)ds$.
Hence \eqref{e2.2} holds.
It follows from Lemma \ref{lem2.1} and \eqref{e2.2} that $T_n$ has
a fixed point $x_n$ in $C_n$. Using $x_n$ and 1 in the place
of $x$ and $\lambda$ in \eqref{e2.2}, we obtain easily
$b_n\leq x_n(t)\leq R, t\in [\frac{1}{n},1]$.
The proof is complete.
\end{proof}
\begin{lemma} \label{lem2.4}
For a fixed $h\in (0,\min\{\frac{1}{2},\eta\})$, suppose
$m_{n,h}=\min\{x_n(t), t\in[h,1]\}$.
Then $m_h=\inf\{m_{n,h}\}>0$.
\end{lemma}
\begin{proof}
Since $x_n(t)\geq b_n>0$, we get $m_h\geq 0$. For any fixed natural
numbers $n$ ($n>[\frac{1}{\eta}]+1$),
let $t_n\in[h,1]$ such that
$ x_n(t_n)=\min\{x_n(t),t\in [h,1]\}$. If $m_h=0$, there exist a
countable set $\{n_i\}$ such that
\begin{equation}
\lim_{n_i\to+\infty}x_{n_i}(t_{n_i})=0 .\label{e2.14}
\end{equation}
So there exist $N$ such that
$x_{n_i}(t_{n_i})\leq \min\{b(t), t\in [\frac{h}{2},1]\}$, $n_i>N$.
Then we have two cases.
Case 1. There exist $n_k\in \{n_i\}, n_k>N$ and
$t^*_{n_k}\in [\frac{h}{2},h]$ such
that $x_{n_k}(t^*_{n_k})\geq x_{n_k}(t_{n_k})$. By the same
argument in Lemma \ref{lem2.3}, we can get
$t'_{n_k}, t''_{n_k}\in [\frac{h}{2},1], t'_{n_k}N$. And so we have
\begin{equation}
\lim_{n_i\to +\infty }x_{n_i}(t)=0, \quad t\in[\frac{h}{2},h]. \label{e2.17}
\end{equation}
On the other hand for any
$t\in[\frac{h}{2},h]$,
\begin{equation}
\begin{aligned}
x_{n_i}(t)&=\frac{2}{h}\int_{\frac{h}{2}}^t(t-\frac{h}{2})(h-s)
f(s,x_{n_i}(s))ds\\
&\quad +\frac{2}{h}\int_t^h(s-\frac{h}{2})(h-t)
f(s,x_{n_i}(s))ds+x_{n_i}(\frac{h}{2})+x_{n_i}(h)\\
&\geq \frac{2}{h}[\int_{\frac{h}{2}}^t(t-\frac{h}{2})(h-s)a(s)ds
+\int_t^h(s-\frac{h}{2})(h-t)a(s)ds]>0,
\end{aligned} \label{e2.18}
\end{equation}
which contradicts \eqref{e2.17}. The proof is complete.
\end{proof}
\section{Main Result}
\begin{theorem} \label{thm3.1}
If {\rm (S1)--(S3)} hold, the three-point boundary-value problem
\eqref{e1.1}-\eqref{e1.2} has at least one positive solution.
\end{theorem}
\begin{proof}
For any natural numbers $n\geq [\frac{1}{\eta}+1]$, it follows
from Lemma \ref{lem2.3} that there exist $x_n\in C_n, b_n\leq x_n\leq R$
satisfying \eqref{e2.1}. Now we divide the proof into three steps.
Step 1. There exist a convergent subsequence of $\{x_n\}$ in (0,1].
For a natural number $k\geq \max\{3,[\frac{1}{\eta}]+1\}$,
it follows from Lemma \ref{lem2.4} that
$00, t\in(0,1]$ by Lemma \ref{lem2.4}.
Step 2. $x(t)$ satisfies \eqref{e1.1}.
Fixed $t\in (0,1]$, we may choose $h\in (0,\min\{\frac{1}{2},\eta\})$
such that $t\in (h,1]$ and
\begin{equation}
\begin{aligned}
x_n(t)&=-\int_h^t(t-s)f(s,x_n(s))ds\\
&\quad +\frac{t-h}{1-\alpha \eta-h(1-\alpha)}
[\int_h^1(1-s)f(s,x_n(s))ds-\alpha \int_h^{\eta}(\eta -s)
f(s,x_n(s))ds]\\
&\quad +x_n(h)+\frac{(t-h)(1-\alpha)}{1-\alpha \eta-h(1-\alpha)}
(b_n-x_n(h)), \quad t\in (h,1].\label{e3.2}
\end{aligned}
\end{equation}
Letting $n\to +\infty$ in \eqref{e3.2}, we have
\begin{equation}
\begin{aligned}
x(t)&=-\int_h^t(t-s)f(s,x(s))ds
+\frac{t-h}{1-\alpha \eta-h(1-\alpha)}\\
&\quad\times \big[\int_h^1(1-s)f(s,x(s))ds
-\alpha \int_h^{\eta}(\eta -s)f(s,x(s))ds\big]\\
&\quad +x(h)+\frac{(t-h)(1-\alpha)}{1-\alpha \eta-h(1-\alpha)}(-x(h)),
\quad t\in (h,1].\label{e3.3}
\end{aligned}
\end{equation}
Differentiating \eqref{e3.3}, we get the desired result.
Step 3. $x(t)$ satisfies \eqref{e1.2}. Let
$$
t_n=\inf \{t:x_n(t)=\|x_n\|, x'_n(t)=0, t\in[\frac{1}{n},1]\},
$$
where $\|x_n\|=\max _{\frac{1}{n}\leq t\leq 1}x_n(t)\leq R$.
Then
$$
t_n\in[\frac{1}{n},1], \quad
x_n(t_n)=\|x_n\|, \quad
x'_n(t_n)=0.
$$
Using $x_n(t), 1$ and $t_n$ in place of $x(t), \lambda$ and
$t'$ in Lemma \ref{lem2.3}, we obtain easily by \eqref{e2.13}
\begin{equation}
\int_{b_n}^{\|x_n\|}\frac{dx}{F(x)}
\leq (1+\frac{\bar{G}(R)}{F(R)})\int_0^{t_n}\int_t^1k(s)\,ds\,dt.
\label{e3.4}
\end{equation}
It follows from \eqref{e3.4} and Lemma \ref{lem2.4} that
$0x_2(\eta)$. That is to say
$z(\eta)>0, 0z(1)>0, \quad t\in(t_1,t_2).
\]
Letting $s(t)=z(t)-z(1)$, we have that $s(t_1)=s(t_2)=0$ and
$s(t)>0, t\in(t_1,t_2)$.
It follows from \eqref{e1.1} and the monotonicity of $f(t,\cdot)$
that $s''(t)=z''(t)\geq 0, t\in (t_1,t_2)$.
An elementary form of the maximum principle implies
$s(t)\leq 0$ for all $t\in (t_1,t_2)$ and hence a contradiction.
Then, $x_1(\eta)=x_2(\eta)$, which also yields that
$x_1(1)=x_2(1)$. That is to say $z(0)=z(\eta)=z(1)=0$.
We next claim that $x_1(t)=x_2(t), t\in(0,\eta)$. In fact,
if it is not true, without loss of generality,
we can get $x_1(t_0)>x_2(t_0)$ for some $t_0\in(0,\eta)$.
Let $t_3=\max\{t\in(0,t_0), z(t)=0\},
t_4=\min\{t\in(t_0,\eta), z(t)=0\}$(note $z(\eta)=0)$).
Then $z(t_3)=z(t_4)=0$ and $z(t)>0, t\in(t_3,t_4)$.
Let $s_1(t)=z_1(t)-z_2(t)$, $t\in[t_3,t_4]$. Then
$s_1(t)>0$ for all $t\in[t_3,t_4]$. On the other hand,
the monotonicity of $f(t,\cdot)$ implies that
$s''_1(t)\geq 0, t\in (t_3,t_4)$.
An elementary form of the maximum principle implies
$s_1(t)\leq 0$ for all $t\in (t_3,t_4)$ and hence a contradiction.
The same argument yields that $x_1(t)=x_2(t), t\in(\eta,1)$.
Hence we get $x_1(t)=x_2(t), t\in[0,1]$. Thus the result is proved.
\end{proof}
\subsection*{Example}
Consider the second order singular three-point boundary-value problem
\begin{gather}
x''(t)+\frac{1}{4}(x^2(t)+\frac{1}{x^2(t)}-\frac{x^3(t)}{t^5}
-\frac{1}{t^2})=0,\quad 0